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If 1 mol of gas is placed into a balloon under standard temperature and pressure (273 K and 1 atm), what volume would the balloon be?

Respuesta :

sebassandin

Answer:

[tex]V=22.4L[/tex]

Explanation:

Hello,

In this case, considering the ideal gas equation:

[tex]PV=nRT[/tex]

It is possible to compute the volume the gas would have for the given STP conditions as:

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{1mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}\\\\V=22.4L[/tex]

Which correspond to the standard volume as well.

Best regards.

CintiaSalazar

Answer:

The volume of the balloon would be 22.386 L

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 1 atm
  • V= ?
  • n= 1 mol
  • R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]
  • T= 273 K

Replacing:

1 atm* V= 1 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex]*273 K

Solving:

[tex]V=\frac{1 mol*0.082\frac{atm*L}{mol*K} *273 K}{1 atm}[/tex]

V=22.386 L

The volume of the balloon would be 22.386 L

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